# Confusion in understanding epsilon delta definition of limit with a discontinuous function

To understand the definition (from Wikipedia)

I have taken a function $$f(x) = \begin{cases} 5 \quad &\text{if x \le 2,} \\ 5 \quad &\text{if x=3,} \\ 5 \quad &\text{if x\ge 4,} \end{cases}$$

I also see that $$\forall \epsilon \gt 0, \exists \delta=2 \gt 0, \forall x \in D, 0 \lt |x-3| \lt \delta \implies |f(x) - 5| \lt \epsilon$$ where $$D = \mathbb{R} \setminus \{(2,3) \cup (3,4)\}$$

With this, it seems the limit for $$f(x)$$ exists at $$x=3$$ even tough there is a big gap around $$x=3$$. Wondering if I'm misinterpreting the definition! Is there any bound on the gap in the neighborhood of $$x=c$$, that I'm not able to gather from the definition of limit.

• Why is $f$ discontinuous? Sep 3, 2020 at 18:03

It seems like you're quoting this passage from Wikipedia. Note the condition stated at the beginning of that paragraph, that $$c$$ has to be a limit point of $$D$$. If it's not (like $$c=3$$ in your example), then it's not meaningful to even begin talking about “$$\lim\limits_{x \to c} f(x)$$”, so the limit doesn't exist in that case.